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ilikephysics2
 2 years ago
Find the first 4 terms of the taylor series of sin x at x = pi/4.
ilikephysics2
 2 years ago
Find the first 4 terms of the taylor series of sin x at x = pi/4.

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ilikephysics2
 2 years ago
Best ResponseYou've already chosen the best response.0The first 4 derivatives are cos(x) then sin(x) then cos(x) and then sin(x)

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.0sin(pi/4) + cos(pi/4) * (xpi/4) sin(pi/4) * (1/2!) (xpi/4)^2 .....

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.0pretty straightforward.. what was your question about?

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.0that's the first 3 terms... you should be able to get the 4th now?

ilikephysics2
 2 years ago
Best ResponseYou've already chosen the best response.0That does not look right though

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.0well... I lied a bit... a few of those terms are zero...

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.0but you can work from them... right?

RadEn
 2 years ago
Best ResponseYou've already chosen the best response.0f(x) = f(x0) + (xx0)*f'(x0)/1! + (xx0)^2*f"(x0)/2! + (xx0)^3*f'"(x0)/3!

RadEn
 2 years ago
Best ResponseYou've already chosen the best response.0with x0 = pi/4, for f(x) = sinx > f(pi/4) = sin(pi/4) = 1/2*sqrt(2) f'(x) = cosx > f'(pi/4) = cos(pi/4) = 1/2*sqrt(2) f"(x) = sinx > f"(pi/4) = sin(pi/4) = 1/2*sqrt(2) f'"(x) = cosx > f'"(pi/4) = cos(pi/4) = 1/2*sqrt(2) now, just subtitute all numbers above to generally formula !!!
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